Uniqueness of a point where the chord coincides with the tangent

Question

Let $F:[0,1] \to [0,\infty)$ be a $C^2$ function satisfying $F(1)=0, F'(1)=0$, which is strictly strictly decreasing on $[0,1]$. Suppose that for some $a \in(0,1)$, $F'' < 0$ on $[0,a)$ and $F'' > 0$ on $(a,1]$.

I am trying to prove that there exist a unique point $s \in (a,1]$ satisfying satisfying $\frac{F(s)-F(0)}{s-0}=F'(s)$.

I have a proof for the existence (see below), but I am having trouble with establishing uniqueness. I vaguely remember that I had an argument for that, but somehow I can't reconstruct what it was exactly.

Proof of existence:

Define $H(s)=\frac{F(s)-F(0)}{s-0}-F'(s)$. Then $H(1)=-F(0)<0$. The concavity of $F$ on $[0,a)$ implies that $H(s) >0$ on $(0,a)$. Indeed $H(s)=F'(\theta(s))$ for some $\theta(s) \in (0,s)$ and $F'$ is decreasing on $(0,a)$. By the intermediate value theorem $H(s)$ must be zero for some $s \in (a,1]$.

---

*I am actually interested in proving this under the weaker assumption that $F \in C^1, F|_{[0,a]},F|_{[a,1]} \in C^2$ but the second derivatives do not agree at $a$.

---

Answer 1

Consider the function $h(x) = F(x) - x F'(x)-F(0)$. Then

• $h(a) > 0$ because $F$ is strictly concave on $[0, a]$,
• $h(1) = -F(0) < 0$,
• $h'(x) = -xF''(x) < 0$ on $(a, 1]$, so that $h$ is strictly decreasing on $[a, 1]$.

This implies that there is a unique point $s \in (a, 1)$ satisfying $$ h(s) = 0 \iff \frac{F(s)-F(0)}{s-0} = F'(s) \, . $$